A set of three cubes are to be glued together so the resulting figure has the minimum surface area. The specific case with edge lengths {1,2,4} is discussed in
Sugar Cubes.
For a general set of three edge lengths {x,y,z}, what is the minimum surface area? (This may be dependent on the relative sizes of x, y ,and z.)
In Sugar Cubes, the largest cube had edges at least as large as the total of the edges of the other two cubes. If this is not the case, then there will be an overhang of the pair of smallest and middle-sized cubes over the edge of the largest. That overhang will be equal to the sum of the two small cubes' edges minus the edge length of the largest cube and adds to the surface area, so we want to minimize this overhang.
WLOG, let the x, y and z of the puzzle be in ascending order.
If x+y <= z, the solution is of the same form as for Sugar Cubes:
6*x^6 + 6*y^2 + 6*z^2 - 4*x^2 - 2*y^2
= 2*x^2 + 4*y^2 + 6*z^2.
The overhang may as well be of the smallest cube (to get the minimal added area), so we get:
6*x^6 + 6*y^2 + 6*z^2 - 4*x^2 - 2*y^2 + 2*x*(x+y-z)
= 2*x^2 + 4*y^2 + 6*z^2 + 2*x*(x+y-z)
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Posted by Charlie
on 2016-12-08 10:28:20 |
proposed solution
